I know that a finite group with a prime number of elements is cyclic and every element in the group is a generator for the group.
Thinking about $(\mathbb{Z}_p^*, \cdot)$ I thought that the order of $(\mathbb{Z}_p^*, \cdot)$ was $p$, so that every element in the group had to be a generator, but then I pondered the fact that I had to remove the 0 and so $(\mathbb{Z}_p^*, \cdot)$ had to have cardinality $p-1$, and so the theorem shouldn't be applied anymore.
This also explained to me the presence in $(\mathbb{Z}_p^*, \cdot)$ groups (lets say $(\mathbb{Z}_{19}^*,\cdot)$) of particular elements which generate the entire group named "primitive roots" (the group generators if I understood well), and that not every element of the group is a primitive root.
Is all that I wrote correct? Did I make the correct reasoning?
You are correct. The order of $\mathbb{Z}_p^*$ is $p-1$, so it is not a priori cyclic. However, it turns out to be a nontrivial fact that it is cyclic; and not every element is a generator, because (for instance) 2 divides $p-1$ and so there is an element of order $2$.
Actually finding a generator is very difficult in full generality. As far as I know, it's still open whether $2$ is a generator mod $p$ for infinitely many $p$.